# Show that the normal at P with x+py=2ap+ap^3 passes through point R(-ap,3a+ap^2)

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### 2 Answers

You should remember that you may prove that a given curve passes through a given point if coordinates of the point verify the equation of curve.

Hence, you should substitute `-ap` for x and `3a+ap^2` for y in the given equation `x+py=2ap+ap^3` such that:

`-ap + p(3a+ap^2) = 2ap + ap^3`

You need to open the brackets to the left side such that:

`-ap + 3ap + ap^3 = 2ap + ap^3`

Combining like terms to the left yields:

`2ap + ap^3 = 2ap + ap^3`

**Hence, since the values of coordinates of the given point check the equation, hence, the normal `x+py=2ap+ap^3` passes through the point `(-ap , 3a+ap^2).` **

x + py = 2ap + ap^3

We form this equation to ax + by + c =0

x + py - (2ap + ap^3) = 0

If above equation passes through the R point , x, y coninate of the R point must complete the above equation.

R (-ap, 3a+ap^2)

x= -ap , y= 3a+ap^2

Now we apply x, y to above equation

x + py - (2ap + ap^3)

= -ap + p(3a+ap^2) - (2ap + ap^3)

= -ap + p(3a) + p(ap^2) - (2ap + ap^3)

= -ap + 3ap + ap^3 - 2ap - ap^3

= 0

Therefor,

x + py = 2ap + ap^3 passes through R point